In a photoelectric experiment a parallel beam of monochromatic light with power of $$200W$$ is incident on a perfectly absorbing cathode of work function $$6.25eV$$. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is $$100$$%. A potential difference of $$500V$$ is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $$F=n$$×$$10$$−$$4N$$ due to the impact of the electrons. The value of n $$is_________$$. Mass of the electron $$me=9$$×$$10$$−$$31kg$$ and $$1.0eV=1.6$$×$$10$$−$$19J$$.

Question

In a photoelectric experiment a parallel beam of monochromatic light with power of $$200W$$ is incident on a perfectly absorbing cathode of work function $$6.25eV$$. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is $$100$$%. A potential difference of $$500V$$ is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $$F=n$$×$$10$$−$$4N$$  due to the impact of the electrons. The value of n $$is_________$$. Mass of the electron $$me=9$$×$$10$$−$$31kg$$ and  $$1.0eV=1.6$$×$$10$$−$$19J$$.

Infinity Learn · Accepted Answer

power $$=$$ nhv           $$n=$$ number of photons per secondSince $$KE$$$$=$$ $$0$$,  $$hv=$$ϕ$$200=n6.25$$×$$1.6$$×$$10$$−$$19$$​joule $$n=2001.6$$×$$10$$−$$19$$×$$6.25$$   As photon just above threshold frequency KEmax is zero and they are accelerated by potential difference of $$500V$$. $$KEf=q$$ΔV       $$P22m=q$$ΔV ⇒$$P=2mq$$ΔV     Since efficiency is $$100$$%, number of electrons ejected $$=$$ number of photons per secondAs photon is completely absorbed force exerted $$F=nmv=np=2006.25$$×$$1.6$$×$$10$$−$$19$$×$$29$$×$$10$$−$$31$$×$$1.6$$×$$10$$−$$19$$×$$500=3$$×$$200$$×$$10$$−$$25$$×$$16006.25$$×$$1.6$$×$$10$$−$$19=2$$×$$406.25$$×$$1.6$$×$$10$$−$$4$$×$$3=24$$×$$10$$−$$4NThus$$, n $$=$$ $$24$$

Photoelectric effect and study of photoelectric effect

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material